Explanation

(b)
For rational roots $b^{2} - 4ac$ must be a perfect square of a rational number and as $a,b,c$ are natural numbers $b^{2} - 4ac$ must be a perfect square of an integer.
$b^{2} - 4ac = I^{2}\ \Rightarrow b^{2} = I^{2} = 4ac$
$\Rightarrow 4ac = (b - I)(b + I)$
$\Rightarrow ac = \frac{b - I}{2} \bullet \frac{b + I}{2}$
$b - I,\ b + I$ are both odd integers or both even integers but $ac$ is an odd integer. So, $b - I$ and $b + I$ must be even integers. $b$ is odd $I$ must be odd. Now, let
$b - I = 2m$, $(m$ odd integer)
$b + I = 2n,\ (n$ odd integer)
$I = (n - m),(n - m$ is an even integer)
So, contradiction $\Rightarrow \ b^{2} - 4ac$ is not a perfect square. So, all $a,b,c$ cannot be odd integers.